Perform the indicated Integration: ∫ex/(1+2ex)dx (but 2 diff. answers??)

[Tonia]

The problem is ∫e^x/(1+2e^x)dx and the steps are:
u = 2e^x + 1----> du/dx = 2e^x
So: ∫e^x/(1+2e^x)dx = 1/2∫1/u du
Plug in solved integrals: 1/2∫1/u du = ln(natural logarithm)(u)/(2)
So then Answer is: ∫e^x/(1+2e^x)dx = ln((2e^x + 1)/(2)) + C
OR Answer is: √1 + 2e^x + C (1 and 2e^x are both under the square root)


These two answers are from difference sources for this one problem. I am teaching myself Integrals. The answer from my textbook is:
ln(natural logarithm)√1 + 2e^x + C (the 1 and 2e^x are both under the square root) and the answer from an automatic Integral Calculator is(I used the integral calculator to help me with some of the steps):
ln((1+2e^x)/(2)) + C
My question is, as a rule, is ln√x equal to ln(x)/(2)?? which means that ln((1+2e^x)/(2)) + C is equal to ln√1 + 2e^x + C??
My other question is: where did the 1/2∫1/u come from and why does it equal ln((u)/(2))??

 

[stapel]

The problem is ∫e^x/(1+2e^x)dx and the steps are:
u = 2e^x + 1----> du/dx = 2e^x
So: ∫e^x/(1+2e^x)dx = 1/2∫1/u du
Plug in solved integrals: 1/2∫1/u du = ln(natural logarithm)(u)/(2)
So then Answer is: ∫e^x/(1+2e^x)dx = ln((2e^x + 1)/(2)) + C
OR Answer is: √1 + 2e^x + C (1 and 2e^x are both under the square root)

These two answers are from difference sources for this one problem. I am teaching myself Integrals. The answer from my textbook is:
ln(natural logarithm)√1 + 2e^x + C (the 1 and 2e^x are both under the square root) and the answer from an automatic Integral Calculator is(I used the integral calculator to help me with some of the steps):
ln((1+2e^x)/(2)) + C

My question is, as a rule, is ln√x equal to ln(x)/(2)?

Why? What did you get when you converted the radical to fractional-power notation, and then applied the log rule for powers inside logs?

My other question is: where did the 1/2∫1/u come from and why does it equal ln((u)/(2))??

They used an obvious u-substitution. But there can be more than one way to do an integral. What u-substitution did you use? :wink:

 

[lookagain]

My question is, as a rule, is ln√x equal to ln(x)/(2)?? \(\displaystyle \ \ \ \ \ \) Yes.

which means that ln((1+2e^x)/(2)) + C is equal to ln √1 + 2e^x + C ** ??
My other question is: where did the 1/2∫1/u come from and why does it equal ln((u)/(2))??

** That needs grouping symbols:

ln √(1 + 2e^x) + C

- - - - - - - - - - -- - - -- - - -- - - - --- - - - - --- - - - - -- -- -